Problem 47 - College Algebra and Calculus: An Applied Approach

Question

Optimal Area The perimeter of a rectangle is 200 feet. Let \(x\) represent the width of the rectangle. Write a quadratic function for the area of the rectangle in terms of its width. Find the vertex of the graph of the quadratic function and interpret its meaning in the context of the problem.

Step-by-Step Solution

Answer

The quadratic function for the area is \(100x - x^2\). The vertex of the function is at \(x = 50\), meaning that the optimal width (and length) for the greatest area of the rectangle is 50 feet.

Step 1: Find the Length

Since the perimeter of a rectangle is \(2*(length + width)\) and the width is represented by \(x\), let's denote the length by \(y\). Therefore, \(200 = 2*(x + y)\). Solving this equation for \(y\), we find \(y = 100 - x\).

Step 2: Write a Quadratic Function for the Area

The area of a rectangle is the product of its length and its width. Substituting the expression we found for \(y\) (the length) into this formula, we get: \(Area = x * (100 - x) = 100x - x^2\). This is our quadratic function.

Step 3: Find the Vertex of the Quadratic Function

To find the maximum area (which is the vertex of the quadratic function graph), we need the formula for the vertex of a quadratic function: \(h = -b/(2a)\). In our quadratic function \(100x - x^2\), \(a = -1, b = 100\), so the vertex \(h = -100 /(2*(-1)) = 50\).

Step 4: Interpret the Vertex

The vertex (\(h\)) represents the value of \(x\) (the width of the rectangle) that maximizes the area of the rectangle. Thus, for a perimeter of 200 feet, the maximum rectangular area is achieved when the width (and thus also the length, since it's a square) is 50 feet.

Key Concepts

Vertex of a Quadratic

In mathematics, a quadratic function is generally expressed in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The graph of a quadratic function forms a parabola, which can either open upwards or downwards depending on the sign of \( a \). A key feature of this parabola is its vertex, which is the highest or lowest point of the graph.
The vertex provides valuable information about the function, particularly when trying to find maximum or minimum values. The formula to find the \( x \)-coordinate of the vertex in the function \( ax^2 + bx + c \) is \( h = -\frac{b}{2a} \).
For our function \( 100x - x^2 \), the vertex is where the maximum area occurs. Here, \( a = -1 \) and \( b = 100 \), so \( h = -\frac{100}{2(-1)} = 50 \). Thus, the vertex \( x = 50 \) tells us that at a width of 50 feet, the area reaches its maximum. This particular solution points out that both the length and width should be equal when the perimeter constraint is applied.

Maximum Area

When dealing with a geometric problem like optimizing the area of a rectangle, we can use quadratic functions to help us find the solution. The area \( A \) of a rectangle can be expressed in terms of its width \( x \) and length \( y \), typically as \( A = x \times y \).
In the given problem, the perimeter restricts the dimensions of the rectangle. Its quadratic nature comes from expressing the area in terms of one variable by substituting for the other, based on known constraints.

• Perimeter of 200 means \( 2(x + y) = 200 \).
• Solving gives \( y = 100 - x \).

So the area becomes \( x \times (100 - x) \), transforming into \( 100x - x^2 \). This equation is a downward-opening parabola, signaling a definite maximum point, which is found using the vertex.
The vertex in this case shows that dividing the perimeter equally (into a length and width both of 50) results in maximum area—as it turns out, a square gives the most enclosed space for the least amount of perimeter.

Perimeter of Rectangle

The perimeter of a rectangle plays a crucial role in problems involving optimization of area. Understanding perimeter is fundamental to manipulating the relationship between length and width. Defined simply, the perimeter \( P \) is the sum of all the sides of the rectangle, expressed as \( 2(l + w) \), where \( l \) is the length and \( w \) is the width.
In situations where we have a fixed perimeter, it limits the number of valid dimensions for length and width.

• If \( x \) represents the width, length \( y \) is determined by \( y = \frac{P}{2} - x \).

This substitution allows us to express area in a way conducive to finding maximum or minimum by transforming into a quadratic equation. When setting up such a problem, always ensure to double-check units and logical context - like making sure dimensions don't exceed what the perimeter allows. When both \( x \) and \( y \) equal 50, the shape is a square, which explains why this configuration is frequently optimal for maximum area.